Let us assume: $f(1) = x.$
Then, $f(2) = f(2/2) = f(1) = x$
$ f(3) = f(3+5) = f(8) = f(8/2) = f(4/2) = f(2/1) = f(1) = x.$
Similarly, $f(4) = x$
$f(5) = f(5+5) = f(10/2) = f(5) = y.$
So, it will have two values. All multiples of $5$ will have value $y$ and others will have value $x.$